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michael glamore March 14, 2006 Equilibrium and Le Chatelier's Principle Britany Held Purpose: This laboratory experiment is able to test a student's knowledge of stoichiometry and promote a general understanding of the Le Chatlier's Principle. The general operations of the lab proceed four different parts, each with their own specific methods and problems. The majority of the lab is simply following directions with little actual thinking or consideration but the lab report requires students to take additional steps in calculation. Formulas needed for experimentation: (Pressure)(Volume) = (Number of Moles)(Universal Gas Constant)(Temperature) *Note: If units are atm, L, mol, and K respectively, then Universal Gas Constant is equal to .08206 Molarity = Moles per Liter K = Ae-Ea / RT Rate = Ae-Ea/RT[I-]x[S2O82-]y Observations: A) Supersaturation and the Effect of Temperature on Equilibrium The experiment begins with thorough cleaning of test tubes and the covering of the inner tube with paper at to keep the reaction from occurring at any other level besides the bottom of the tube. These two test tubes contain Sodium acetate with the addition of water. Then, after heating and separating the crystals, CoCl2*6H2O and NaC2H3O2*3H2O are each added to a test tube. The Sodium Acetate reacts in the following way: NaC2H3O2*3H2O(s) <--> Na+(aq) + C2H2O2-(aq) + 3H2O(l) Producing a white substance that reacts instantly forming an almost immediate equilibrium. The CoCl2*6H2O reacts in the following way: CoCl2*6H2O(s) + NaC2H3O2*3H2O(s) <--> NaCl(aq) + CoC2H2O2(aq) + 3H2O(l) This equilibrium takes significantly longer to react which eventually takes the pinkish color of CoC2H2O2(aq) but takes significantly longer to do so. B) Effect of Concentration on Equilibrium and the Calculation of K SbCl3(aq) + H20(l) <--> SbOCl(s) + 2HCl(aq) where the constant k is equal to: [HCL]2 k = [SbCl3] B-1) Adding water to the SbCl3 until equilibrium is reached should allow us to determine a correct value of k. The solution of acidic antimony trichloride solution (0.5M SbCl3 in 6M HCl) of 5 ml. It took almost exactly 13.8 ml to create a solution that would not go back to its clear form but rather stay in its white precipitate (SbOCl). Using a nice little trick with the total volume (19ml) and the overall molarities of HCl -6M and .5M SbCl3 the overall k can be calculated. 5 / 1000 * .5 = .0025 SbCl3 5 / 1000 * 6 = .03 HCl 5ml + 13.8ml = 18.8ml / 1000 = .0188 .03 / .0188 = 1.5957M HCL .0025 / .0188 = .13297M SbCl3 (1.5957)2 / (.13297) = 19.17 = k B-2) The hydrochloric acid breaks up the precipitate SbOCl(s) to make a solid white solution. B-3) The experiment in part 1 was carefully repeated to yield similar results this time almost exactly 13.8 ml as an average was concluded to be the appropriate volume of water necessary to add to the substance. K is still equal to 19.17. C) The effects of concentration, Solvent, and Temperature on Equilibrium [Co(H20)6]2+ + 4Cl- <--> [CoCl4]2- + 2H2O C-1) The CoCl2*6H2O is a pinkish red while Co(NO3)2*6H2O is a bit darker more towards the red end of the spectrum CoCl2*6H2O(s) <--> [Co(H20)6]2+(aq) + Cl22-(aq) Co(NO3)2*6H2O(s) <--> [Co(H20)6]2+(aq) + (NO3)22-(aq) C-2) With the addition of various chemicals the colors changed dramatically. The addition of hydrochloride caused a purple and blue solution while the ethanol and HCl quickly turned after just adding 6 drops. This adds to the possibility that ethanol may be a possible catalyst. C-3) [Co(H20)6]2+ + 4Cl- <--> [CoCl4]2- + 2H2O By varying temperature in this previous equation we are able to change the color and therefore the concentration of the two cobalt complexes. At higher temperatures there is more energy pushing the potential energy higher and putting the equation at the right side of the equation. Lower temperatures have less energy and create a very pink solution at the left side of the equation. Room temperature supports a more neutral violet, a seeming mix of the two compounds. D) The Effect of Solubility on the Completeness of Chemical Reactions This experiment involves the manipulation of a lead ion to various salts, using supernatants and selective precipitates in different ways. It begins with this mixture: Pb(NO3)(aq) + NaCl(aq) <--> PbCl(s) + NaNO3(aq) The white sold formed at the top of the container is Lead Chloride, centrifuging this multiple times allows the solid to be fully separated from the other substances. The addition of Na2CO3 allows for this chemical equation: PbCl(s) + Na2CO3(aq) <--> PbCO3(s) + 2Na-(aq) This addition allows for the creation of solid Lead Carbonate as a precipitate. Centrifuging the two parts fully separates them into two parts. Taking the supernatant Sodium in this reaction: 2Na-(aq) + Na2S(aq) <--> 4Na-(aq) + S(s) Getting rid of the precipitate sodium, we are left with pure sulfur which creates a brownish mixture. 2HNO3(aq) + S(s) <--> H2S(g) + 2NO3(aq) This creates an egg smell that comes in bubbles off as it is mixed. With the original precipitate, Lead Carbonate, we add HNO 3 in the following reaction: PbCO3(s) + 2HNO3(aq) <--> Pb(NO3)2(s) + CO2(g) + H2O(l) This creates a colorless gas with no smell, carbon dioxide. Conclusion: This experiment worked very well as it allowed us to fully understand the consequences of equilibrium in almost all cases in almost every experiment. Even water is in constant equilibrium. Interesting enough, the concepts will come in handy in working with any substance in the future.

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